As mentioned, this is not true. However, a related statement is true: If $f$ is uniformly continuous and bounded, then $f^2$ is uniformly continuous (and bounded). (Hint to prove this: $|f(x)^2 - f(y)^2| = |f(x)+f(y)| |f(x)-f(y)|$). Indeed, if $f,g$ are both uniformly continuous and bounded, then so is $fg$. Thus any finite product of uniformly continuous bounded functions is another such; in particular, if $f$ is, then so is $f^n$ for any $n$.

A fancy version of this statement is that the set $C_u(\mathbb{R})$ of all uniformly continuous bounded functions on $\mathbb{R}$ is a $C^*$-algebra.